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2015-07-12T11:42:38+05:30

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A³ - b³ + 1 + 3ab
a³ – b³ + 1 + 3ab = (a – b)(a² + ab + b²) + 1 + 3ab 

 adding 1 inside the first brackets and subtract 1 times the second brackets to leave the expression unchanged in value 

= (a – b + 1)(a² + ab + b²) –1(a² + ab + b²) + 1 + 3ab 
= (a – b + 1)(a² + ab + b²) – a² – ab – b² + 1 + 3ab 
= (a – b + 1)(a² + ab + b²) – (a² + ab + b² – 1 – 3ab) 
= (a – b + 1)(a² + ab + b²) – (a² + b² – 1 – 2ab) 

as (a – b)² ≡ a² – 2ab + b² 

 = (a – b + 1)(a² + ab + b²) – ((a – b)² – 1) 


Now see the difference of two squares 
((a – b)² – 1) = ((a – b)² – 1²) = (a – b + 1)(a – b – 1) 

Hence 
 = (a – b + 1)(a² + ab + b²) – (a – b + 1)(a – b – 1) 

 = (a – b + 1)[(a² + ab + b²) – (a – b – 1)] 

= (a – b + 1)(a² + ab + b² – a + b + 1)ANSWER

6 4 6
hope this helps
Which standard are you?
there is no direct algebraic identity to factorise that
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